A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220
1 answer:
First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force,
. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
Now we can use the following relationship to find the distance covered by the skier before stopping, S:
where
is the final speed of the skier and 
is the initial speed. Substituting numbers, we find:
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force,
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
Now we can use the following relationship to find the distance covered by the skier before stopping, S:
where
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Answer:
V = 10.88 m/s
Explanation:
V_i =initial velocity = 0m/s
a= acceleration= gsinθ-cosθ
putting values we get
a= 9.8sin25-0.2cos25= 2.4 m/s^2
v_f= final velocity and d= displacement along the inclined plane = 10.4 m
using the equation
v_f= 7.04 m/s
let the speed just before she lands be "V"
using conservation of energy
KE + PE at the edge of cliff = KE at bottom of cliff
(0.5) m V_f^2 + mgh = (0.5) m V^2
V^2 = V_f^2 + 2gh
V^2 = 7.04^2 + 2 x 9.8 x 3.5
V = 10.88 m/s
Answer:
a) The acceleration experimented by the passenger when she passes through the lowest point of her circular motion is: ,
.
b) Hence, the acceleration experimented by the passenger when she passes through the highest point of her circular motion is: ,
.
c) The Ferris wheel takes 14.646 seconds to make a revolution.
Explanation:
a) An object that rotates at constant angular velocity reports a centripetal acceleration and no tangential acceleration. When passenger passes through the lowest point in her circular motion, centripetal acceleration goes up to the center.
In addition, centripetal acceleration is determined by the following expression:
(Eq. 1)
Where:
- Centripetal acceleration, measured in meters per square second.
- Linear speed, measured in meters per second.
- Radius of the Ferris wheel, measured in meters.
If we know that and
, the magnitude of radial acceleration is:
The acceleration experimented by the passenger when she passes through the lowest point of her circular motion is: ,
.
b) In the highest point the magnitude of radial acceleration is the same but direction is the opposed to that at lowest point. That is, centripetal acceleration goes down to the center.
Hence, the acceleration experimented by the passenger when she passes through the highest point of her circular motion is: ,
.
c) At first we need to calculate the angular velocity of the Ferris wheel (), measured in radians per second, by using the following expression:
(Eq. 2)
If we know that and
, then the angular velocity of the Ferris wheel is:
Now we proceed to obtain the period of the Ferris wheel (), measured in seconds, which is the time needed by that wheel to make on revolution:
(Eq. 3)
Where is the angular velocity of the Ferris wheel, measured in radians per second.
If we get that , then:
The Ferris wheel takes 14.646 seconds to make a revolution.